Infinite Resistor Problem - Confirmation

Cute!

So what's the xkcd resistance -- a knight's move?

Tim

-- Deep Friar: a very philosophical monk. Website:

Oops, plain forgot to include that. Just uploaded a new version with it included. Basically, 0.7855 was the lowest value I measured, and the expected value is

Dave.

Neet.

Hey, (8 - pi) / 2pi = 8/2pi - pi/2pi = 4/pi - 1/2. So it's double the diagonal value, but minus the sideways value... coincidence? Now I wonder if there is a general (x, y) solution with factors of 1/2 and pi!

'Ya know, you got about twice the error of the other moves. And it's about twice the distance -- the others were approximately 1 or 2 resistors across, but this is more like 3. So the majority of the current is flowing over a larger area, and your grid is effectively about half the size it was on the smaller measurements. The other measurements still came out higher than predicted, by an amount roughly proportional to the grid. Going by just "1%", it's not statistically significant, but the numbers are suspicious enough that I wonder if your resistors are more accurate than a flat +/-1% distribution.

So what's the error? You should have a fairly well defined (although still approximate) correction factor to go between finite and infinite cases. That correction factor will always be between 0 and 1, because a larger grid can only have less resistance. Obviously, it's related to the gross area of the grid and proximity to the edges, which you have illustrated. Interesting to contemplate what functions would approximate it -- a wave of the hand might simply say, well the grid is N wide and the measurement is n wide, so the error is roughly n/N (or maybe (n/N)^2, or some other power, since you're getting better than 1/14th accuracy, after all). Likewise, the error might range between "fairly close" in the center, to "closer to R" towards the edges, so it might have a (|x| + |y|), or maybe |x*y|, or even x^2 * y^2 sort of factor to it.

It would also be interesting to run a Monte-Carlo on this grid to see what real variations do. Precision across the whole grid goes as N^(-1/2) (for number of resistors N), but most of the grid doesn't matter much, so it might be closer to N^(-1/4) (4th root!) instead. I don't like the idea of analyzing random values in a 500-dimensional space though...

Tim

Coincidences like that don't happen that often in engineering!, I'd almost expect a neat solution like that.

Could well be. Resident mathematicians get to work!

I believe they are much better than 1%. I measured quite a number of them before I started and they were all very much under 1%, like well under 0.5% or so if I remember correctly. I have some left over from the box so I can perform some post construction confirmation...

Perhaps I should perform some systematic single resistor measurements on the central part of the grid to give a better base accuracy to work with for the diagonal measurements. At the moment my numbers are just from memory poking around the grid, I didn't record them.

Indeed. As you go further out the tolerance in each resistor would become irrelevant pretty quickly. In the single resistor case, that single resistance will dominate the error for that reading, so it's possible to post-analyse and even "calibrate" the actual grid this way for the diagonal measurements.

Building the thing has really made the whole problem a lot more interesting!

Dave.

Some folks don't do idiots' blogs.

If the result or the proof so damned elaborate that you couldn't post it here, it must be incorrect.

Only if the truck is approaching at 70mph (up two,over one).

Any slower and all you get is a pawn's move, and a flat pawn.

That's funny!

The (8-pi)/(2pi) comes from a closed solution of the integrals done on paper with pencil, not a software program. There are at least three usefully different ways to approach the solution. The one I used was the Fourier approach, where a constant factor of (1/2pi)^D appears in the front and D is 2 for a 2-dimensional grid. [In other words, there is always a common factor of 1/(4*pi^2) in all answers for a 2D grid, before worrying about the result of the integrals/sums.]

The program was set up to only use odd numbers for the width and height (so that [0,0] could be directly in the center.) The grid I used earlier was 121x121 and I had selected points [0,0] and [1,2] in a grid going from -60 to +60 on each side. (I had been playing with different sizes and when I wrote, I only pointed out that using a grid that spans 30 or more seems to work pretty well -- not that I was actually using a 30x30 in producing those values.)

That isn't exactly centered on the grid. Centering it on that large grid provided (today) a value of 0.7730 instead of the .77284 that I'd reported earlier.

Using a 15x15 node matrix (which is exactly what you'd get with a

In these calculations, the software is using the relaxation algorithm as I earlier mentioned. It is NOT using a Spice simulation with an exact matrix solution to a finite resistor problem, which likely will provide a different result.

In fact, just a second and I'll try it in LTSpice.

Ah. LTSpice provides 0.78561 Ohms for a 14x14 matrix of resistors, with the battery placed so that the mid-point node is straddled. You mentioned seeing nothing less than 0.7855, which is darned close to what Spice says, too.

Take note that the relaxation algorithm on a finite grid is NOT the same as an exact matrix solution that Spice uses. It is just an easy way to write some simple code that is very effective in a broad set of circumstances. In fact, the relaxation algorithm can be easily applied to electric field potentials in a vaccum with varying types of electrode shapes and enclosures and so on. It's got its own areas where it is very effective (Laplace solutions.)

In fact, now that I've mentioned this I was easily able to find this site:

Which has a nice spreadsheet you can use for exactly what I'm doing with the tiny bit of code I wrote, earlier. I modified both the first and second spreadsheet they offer to change the electrode and add a cell that easily displays the same results that my program provides.

If you want my thoughts on changing those spreadsheets, I'll gladly write them up. (It's not much, but there are some details to worry about.)

Jon

Just like your assertion that the standard 0.5R solution is wrong, huh? We are still waiting for your brilliant answer and proof...

For those that don't want to watch my video blog, I actually built a resistor network for some fun:

Dave.

Next time use 0201 SMD. ;-)

As i remember the standard 0.5R solution, the argument is: current into (any) node breaks to 4 equal values, and current out of (any other) node collects from 4 equal notes; "therefore" the resistance is 0.25 (in) + 0.25 (out) for a total of 0.5 ohms. The obvious problem is that the resistance will be larger for widely-spaces probes than for closely spaced probes. I have a sheet of carbon-loaded conductive foam from Jameco, and the measured resistance near the center does indeed follow that assertion. Obviously, the resistance on the diagonal of a given square would be less than the 0.5 ohms given if one used only the 4 resistors of the square. And obviously, resistance across a 2,000 resistor (exact) diagonal would be less than 1K. But in both of these cases, the resistance would not be half of those simplistic values.

Why not 32,767 x 32,767? ;-)

Cheers! Rich

Actually if you picked the right size (probably 0805 or so) you could take a standard prototyping perfboard and use the copper rings as the nodes in the resistor grid. In fact, with solder paste and a toaster oven you could do it really quickly!

Spoken like an owner of a quad-core server with 64gig of ram and a few day's cpu time to waste. ;)

Jon

If someone can prepare a 100x100 for about $10-$20, I'll take one! :) (20200 resistors and 40400 pads -- a walk in the park, eh?)

What I did instead of this was to just generate the surface plot in 3D for a 121x121 node grid (120x120 resistors.) Beautiful to see after the simple Laplacian operator does its work on it.

However, I'd go for a board, too. What does 20200 0201 1k resistors cost with 40400 pads, soldered and ready? :)

Jon

I'll supply the resistors (1K, 0603s) if you do everything else. Where do you want them drop shipped?

You place four pads together, and reduce the pad count. 45° angle grid.

I like that, if they'll self-seat okay.

Jon

I can't even see them. The darned things swirl onto the floor from the vortices of a buzzing gnat. Like finding particles of talcum powder. Might as well be, to my soldering station. I have a hard enough time with sot23-6 micros, already. And at least I can see those.

Jon